limit as x approaches 0 of (tan(2x))/(5x)

Lösung

x \to 0 lim (\frac{tan ( 2 x )}{5 x})

\frac{2}{5}

Dezimale

0.4

Schritte zur Lösung

x \to 0 lim (\frac{tan ( 2 x )}{5 x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{5} \cdot x \to 0 lim (\frac{tan ( 2 x )}{x})

Wende das L'Hopital Theorem an

= \frac{1}{5} \cdot x \to 0 lim (\frac{sec ^{2} ( 2 x ) \cdot 2}{1})

Setze den Wert

x = 0

ein

= \frac{1}{5} \cdot \frac{sec ^{2} ( 2 \cdot 0 ) \cdot 2}{1}

Vereinfache

\frac{1}{5} \cdot \frac{sec ^{2} ( 2 \cdot 0 ) \cdot 2}{1} : \frac{2}{5}

= \frac{2}{5}

\int \frac{4 sin ( x )}{( 1 + cos ( x ) ) ^{2}} d x

y^{^{'}} = \frac{e ^{x} - e ^{y}}{1 + e ^{x}}

\int \frac{sin ( 2 x )}{12 + cos ^{2} ( x )} d x

d er i v a t i v e arccos (∣ x + 1 ∣)

d er i v a t i v e f (x) = 2 x^{2} - 5 x + 3, at a = 2